Solve for $x$ : $2x^2 - 16x + 32 = 0$
Solution: Dividing both sides by $2$ gives: $ x^2 {-8}x + {16} = 0 $ The coefficient on the $x$ term is $-8$ and the constant term is $16$ , so we need to find two numbers that add up to $-8$ and multiply to $16$ The number $-4$ used twice satisfies both conditions: $ {-4} + {-4} = {-8} $ $ {-4} \times {-4} = {16} $ So $(x - {4})^2 = 0$ $x - 4 = 0$ Thus, $x = 4$ is the solution.